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bulgar [2K]
3 years ago
6

PLS HELP DUE TM WILL MARK BRANLIEST !!!

Mathematics
1 answer:
kolezko [41]3 years ago
8 0

Answer:

Step-by-step explanation:

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2(r+3)=<br><br> Can u please help me solve this
AnnZ [28]

Answer:

2r + 6

Step-by-step explanation:

have a good day :)

4 0
2 years ago
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WILL MARK BRAINLIEST
Debora [2.8K]

460 < 2x + 10 and 2x + 10 < 660

Solve x in each equation:

Rewrite so X is on the left side:

2x +10 >460

Subtract 10 from each side:

2x > 450

Divide both sides by 2:

x > 450 /2

x > 225

2x +10 <660

Subtract 10 from both sides:

2x < 650

Divide both sides by 2:

x < 650 / 2

x < 325

Combine the answer: 225 < x < 325

This means her carb intake needs to be more than 225 and less than 325 grams.

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3 years ago
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The height in feet above the water of a diver can be modeled by ℎ() = −16*2 − 16 + 192, where t is the time in seconds after the
Maurinko [17]

h(t) is the height of the diver in feet above the water

Now it asks about the time at which diver reaches the water

When diver reaches the water , the height of diver from water should be zero

So we plug h(t) =0

So

-16t^2 -16 t +192 =0

divide whole equation by -16

t^2 +t -12 =0

we can now factor the quadratic equation

So we get

(t+4)(t-3) =0

plug each factor equal to zero and solve for t

t+4 =0 and t-3 =0

So t=-4 and t=3

Now time cannot be negative , So t=3

So time = 3 seconds

It takes 3 seconds for the diver to reach the water

3 0
3 years ago
DONT IGNORE PLZ HELP ME!!
aleksklad [387]
A = base side=3

b= base side=5

c = base side=3

h= height=7-4=3

l= lenght=6

w= width=5

h= height=4

V= 1/4h√(-a⁴+2(ab)²+2(ac)²-b⁴+2(bc)²-c⁴) + l w h 

V=12.44+ 120=132.44 unit³

7 0
3 years ago
Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile
mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships  is  135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC =  402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40'      [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Taking last two ratio,

\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Putting the value of BC , AC ,∠A,∠B

\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}

\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}

         ≈734.106 miles

Therefore they are 734.106 miles apart.

3 0
3 years ago
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